The distance of the point (7, -3, -4) from th | vedclass

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(C) The equation of the plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equat

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$5 \sqrt{2}$

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(C) The equation of the plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equation:$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}ight| = 0$Substituting the points $(2, -3, 1)$,$(-1, 1, -2)$,and $(3, -4, 2)$:$\left|\begin{array}{ccc} x-2 & y+3 & z-1 \ -1-2 & 1-(-3) & -2-1 \ 3-2 & -4-(-3) & 2-1 \end{array}ight| = 0$$\left|\begin{array}{ccc} x-2 & y+3 & z-1 \ -3 & 4 & -3 \ 1 & -1 & 1 \end{array}ight| = 0$Expanding the determinant:$(x-2)(4 - 3) - (y+3)(-3 + 3) + (z-1)(3 - 4) = 0$$(x-2)(1) - (y+3)(0) + (z-1)(-1) = 0$$x - 2 - z + 1 = 0$$x - z - 1 = 0$The distance $d$ of a point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For the point $(7, -3, -4)$ and plane $x - z - 1 = 0$:$d = \frac{|7 - (-4) - 1|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5 \sqrt{2}$.

The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$,$(-1, 1, -2)$ and $(3, -4, 2)$ is:

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